# Make 2019 with single digits

This puzzle is inspired by a puzzle by @Andrew, where the goal is to make 1998 from a minimal amount of 8s. This is a variation of the puzzle, where the goal is to make 2019 using a minimal amount of single digits (0s, 1s, 2s, etc.)

More formal description (to remove any disambiguations and loopholes):
You need to make a (so-called) valid $$i$$-expression ($$i$$ is a digit from 0 to 9) which evaluates to 2019 under the common sense of the characters being included into it, and has the minimum number of digits (not counting all other characters) in it (the count is separate for each $$i$$), according to the following rules (which determine what an $$i$$-expression is):

• A string of one or more identical digits $$ii\dots i$$ is a valid $$i$$-expression (i.e. $$2$$, $$22$$ and $$22222$$ are valid $$2$$-expressions).
• If $$A$$ and $$B$$ are valid $$i$$-expressions, so are $$A+B$$, $$A-B$$, $$A\times B$$ , $$\frac A B$$, $$A^B$$ and $$\sqrt[A]B$$.
• If A is valid $$i$$-expression, so are $$(A)$$, $$(A)!$$ (factorial - note the parentheses, to explicitly disallow the multifactorials) and $$\sqrt A$$ (here, 2 is implied).

Example (a "5-expression"): $$(5\times(5+5)-5)^{\frac{5+5}{5}}-5-\frac55=(5\times10-5)^\frac{10}5-5-1=45^2-6=2025-6=2019\ (10\ \mathrm{fives})$$

Clarification: The ultimate aim is to find a minimum number for each kind of digit (i.e. a solution with minumum number of zeroes, ones, etc.), with a total of 10 different solutions. Of course, this does not invalidate the existing answers - providing a single solution is already great.

• Should make this an annual puzzle!
Sep 12, 2019 at 9:42
• @Adam Yes, it's a great idea. Sep 12, 2019 at 10:38
• According to this consensus on meta, these types of questions must have a provably optimized answer. It may be the case that the only valid answers are brute-force computer searches. (It may also be too broad, because it asks ten distinct questions.) To answerers: Answers must have a proof of optimality. An answer without a proof of optimality is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 3:54
• If you'd like to discuss this puzzle without knowing whether your solution is optimal, you can do that in the comments. You can also make a chat room for this particular puzzle, and propose solutions there.
– Deusovi
Sep 13, 2019 at 4:06
• @Deusovi is the issue with unproven answers not in the question? If a provable answer is required, I argue that OP should have a proven answer ready, or at least know it is provable. Proving the questions asked in this post is non-trivial and I think fits better in the Mathematics stack exchange. I acknowledge the meta post and I don't want to open that discussion again, but how I read it, the question should not be open ended (rather provable). That forces the answers to come with a proof. Sep 13, 2019 at 6:59

This is an agglomeration of all existing answers.

If you have a more optimal solution than presented above, please replace the answer with your one, change the number of digits used, and remember to change the link to your profile's.

If your solution alters the status of concatenating, please change that too (i.e. with -> without, or without -> with).

Anyone can edit as this is now a community wiki.

# Zero

$$(0! + 0! + 0!)!^{0! + 0! + 0! + 0!} + ((0! + 0! + 0!)!)! + 0! + 0! + 0!$$

# One

$$(1+1)^{11}-(1+1+1)(11-1)+1$$

# Two

$$\sqrt{2^{22}}-2^2!-2-2-\dfrac22$$

# Three

$$3!\sqrt{3!^{3!}}+(3!)!+3$$

# Four

$$\dfrac{\sqrt{\sqrt{4^{4!}}}}{\sqrt{4}} - 4! - 4 - \dfrac{4}{4}$$

# Five

• Using eight 5s with concatenation (@steffen)

$$5 - \frac{5555}{5} + 5^5$$

# Six

$$6\sqrt{6^6}+6!+\dfrac{6+6+6}6$$

# Seven

$$7\left(\dfrac{7!}{7+7}-77+7\right)-\dfrac{77}7$$

# Eight

$$\dfrac{8!}{(\sqrt{8+8})!-\sqrt{8+8}}+\dfrac{(\sqrt{8+8})!}{8}$$

# Nine

$$(\sqrt9\,!)\sqrt{(\sqrt9\,!)^{\sqrt9\,!}}+(\sqrt9\,!)!+\sqrt9$$

• Your $7$s solution betters mine! Sep 12, 2019 at 18:39
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01
• If... without concatenation is better than with, should we replace the without then? Sep 13, 2019 at 13:29
• I mean, without concatenation is the subset of with right? Because without concatenation is more constrained thus with concatenation will be always better than without. (Hence, I suggest if without concatenation is better then we can replace with by without, cmiiw) Sep 13, 2019 at 13:34
• You can use the factorial of the cardinality of an empty set |{}|! = 1, and from that generate any number using no digits!
– user62505
Sep 13, 2019 at 16:27

My own (very trivial) solution with 0s (using 21 of them - so expecting to be beaten by somebody):

$$(0!+0!)^{0!+0!+0!+0!+0!+0!+0!+0!+0!+0!+0!}-(0!+0!+0!)^{0!+0!+0!}-0!-0!\\=2^{11}-3^3-1-1\\=2048-27-1-1\\=2019$$

• The long chain of 0! in the first power could be simplified to (0!+0!+0!)^(0!+0!)+0!+0! . What a monster this will end out being!
Sep 12, 2019 at 11:08
• Please enlighten me and anyone else who doesn't have a degree in mathematics: how the hell do you make a string of 0s mean anything else than 0? I'd really like to understand this one.
– Elmy
Sep 12, 2019 at 17:33
• @Elmy by the magic of factorials! The definition of 0! is 0! = 1
Sep 12, 2019 at 17:46
• @Adam Thanks. But couldn't you shorten the first part to (0!+0!)^(0!0!) instead of adding 0! eleven times? The rules say that ii is a valid 2 digit number, equaling 11 in your case.
– Elmy
Sep 12, 2019 at 20:04
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:00

After an exhaustive search (by hand) I've found a really good solution for 0 using 14 digits!

$$(0!+0!+0!)\times(((0!+0!+0!)!)!-((0!+0!+0!)!+0!))-((0!+0!+0!)!-0!)!$$
$$=3\times((3!)!-(3!+1))-(3!-1)!$$
$$=3\times(6!-(6+1))-(6-1)!$$
$$=3\times(720-7)-5!$$
$$=3\times(713)-120$$
$$=2139-120$$
$$=2019$$

Proof for optimisation

To make this proof less convoluted (and to help in future similar problems) this is a dictionary of the construction of all numbers from 1-10. [Number ($$i$$ count): Expression]

1 (1): $$0!$$
2 (2): $$1+1$$
3 (3): $$2+1$$
4 (4): $$3+1$$ or $$2^2$$
5 (4): $$3!-1$$
6 (3): $$3!$$
7 (4): $$6+1$$
8 (5): $$2^3$$ or $$6+2$$
9 (5): $$3^2$$
10 (6): $$9+1$$ or $$2\times5$$

Proof that it isn't possible to reach 2019 using two (1-10) number $$i$$-expressions, so at least more than 8 digits are required

The only way to reach 2019 using two $$i$$-expressions is to use an expression either of the form $$A \times B$$, $$A^B$$ or $$A-B$$ ($$\frac{A}{B}$$ and $$A+B$$ are redundant). The only factors of 2019 are 3 and 673 and no factorial can equal 673 so $$A \times B$$ is ruled out and $$A^B$$ is ruled out by default. For $$A-B$$ to bear a solution, $$A \ge 2019$$ however no large number $$B$$ obtainable by a 1-10 number expression exists to constrain possible large $$A$$.

Following off of that proof, we now know the approach to take to get 2019

$$A^B$$ and $$\frac{A}{B}$$ are completely out of the question. From this it is safe to say that either $$3\times673=2019$$, $$A+B$$, $$A-B$$ or $$A=2019^B$$

On the subject of factorials...

No perfect power factorial exists above $$1!$$. Also no expression containing only factorials (above $$1!$$) can equal a prime number due to common factors.

This helps to drastically rule out many expressions

$$A!=2019^B$$ is out. $$673$$ is a prime number so the only way to reach it is by $$A \pm B$$. In the case of $$673$$, $$A \pm B$$ is disprovable by exhaustion for 1-10 numbers since nested factorials are disproved. In a more general case, $$A=2019^B$$ cannot be done if $$C$$ is added either. For it to even bear results it would have to be $$A \pm B = 2019^C$$ or $$A \times B = 2019^C$$. Since $$2019^C=3^C673^C$$ it clearly isn't possible using nested factorials so this can be disproved by trivial exhaustion. A common factor of $$3^C$$ can be brought out of $$A!\pm B!$$ for $$A!,B! > 3!$$ leaving two other factorials which must sum or difference to $$673^C$$ which isn't possible because it is many of the same prime. This is now trivially disprovable by exhaustion! Thus this case generally requires at least 12+ digits since more than three (1-10) number $$i$$-expressions are required. I doubt this will lead anywhere so it is fair to say that it has been disproved! {of course this isn't 100% irrefutable proof since it is possible for $$D$$ to enter the equation however the numbers (1-10) would be highly restrained and I don't want to write a full paper on this}

On the subject of $$3\times673=2019$$ we already know that $$3$$ takes 3 digits so all I have to do is show that $$673$$ cannot be made in a reasonable amount of digits. We know that $$A,B$$ expressions that contain only factorials cannot be prime and the other solutions are trivially disprovable by exhaustion ($$A \pm B$$). This brings it up to 11+ digits. Adding in $$C$$, the cases to be examined are $$A \times B \pm C$$, $$A-B \times C$$ and $$A^B-C$$. The unlisted cases are either clearly logically wrong, already listed and/or trivially disproved by exhaustion. $$A \times B \pm C$$ requires a nested factorial somewhere in order for the numbers to be large enough for a solution. If $$A$$ or $$B$$ is a factorial $$>1!$$ then $$C$$ must be an odd number for the result to be prime. Therefore either all numbers are nested factorials (disproved by previous logic) or only one of them are a nested factorial. Since only one of them can be a nested factorial, this is disprovable by exhaustion (gee this is getting exhausting!). This is extended to $$A-B \times C$$ (which is now very obvious without proof by "..."). For $$A^B-C$$, $$A$$ and $$C$$ cannot both be nested factorials so logically only $$B,C$$ should be examined. $$A$$ must then be odd. Looking at $$A^{B!}-C!$$, the cases where $$A \le C$$ - have a common factor thus disproved. $$A>C$$ restricts nested factorial $$C$$ to $$A$$'s upper bound thus is disprovable by "...". All of this proof comfortably disproves $$3\times673=2019$$ since $$673$$ requires more than three (1-10) numbers and at worse case scenario (unlike my $$A=2019^B$$ proof) is +12 digits but should generally require +15 digits.

• @TheSimpliFire got my $1$s down to 13. Sep 12, 2019 at 16:10
• @WeatherVane Great job by both of you! (woulda been weird if concatenation didn't win in the end)
Sep 12, 2019 at 16:14
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:02
• I know that a better answer was found so "proof of optimisation" was changed to "proof for optimisation"
Sep 14, 2019 at 0:50

The best solution I found so far (painful to say with concatenation...):
With 6 copies of 3

$$( (3!)! - 3!3! ) \times 3 - 33$$

$$= ( 6! - 6*6 ) \times 3 - 33$$
$$= ( 720 - 36 ) \times 3 - 33$$
$$= ( 684 ) \times 3 - 33$$
$$= 2052 - 33$$
$$= 2019$$

• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

Update:

The first offering for (sixteen) $$7$$s based on the $$1$$s solution

$$(\frac{7777 - 777 + 77 - 7}{7}) (\frac{7 + 7}{7}) - (\frac{7}{7}) = 1010 \times 2 - 1$$

I have fourteen $$1$$s

$$(1 + 1)(1111 - 111 + 11) - 1 - 1 - 1 = 2 \times 1011 - 3$$

I have eight $$4$$s

$$44(44 + \sqrt{4}) - 4 - \frac{4}{4} = 44 \times 46 - 5$$

• 13: $(1+1)(1111-111+11-1)-1=2019$ Sep 12, 2019 at 15:30
• @TheSimpliFire good spot. Sep 12, 2019 at 16:05
• It seems that for seven, one would be better off not using concatenation -- unless my attempt (12) for seven is beaten... also note that 2016 is divisible by all integers from 1-9 except 5, which makes 2019 = 2016 + 3 easier... Sep 12, 2019 at 18:34
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

With six nines:

$$((\sqrt 9)!)!\times\sqrt 9 -((9+(\sqrt 9)!)\times9) -(\sqrt 9)!$$
$$=720\times3-(15\times9)-6$$
$$=2160-135-6$$
$$=2019$$

• I think you can only find the square root if you have the number 2 to spare
Sep 12, 2019 at 10:37
• @Adam No, you can use the square root symbol without the 2, so the solution is valid. Sep 12, 2019 at 10:40
• Do you see 2 written? No? Then you have no reason to complain. Sep 12, 2019 at 16:23
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

For $$4$$, we can use seven digits:

$$\sqrt{\sqrt{\sqrt{4^{44}}}} - ((4-4)! + 4! + 4)$$

• So next year, only five digits are needed :p Sep 12, 2019 at 11:50
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

Ten 5s (different from @TheSimpliFire and OP)

$$(5*5-5)*((5*5-5)*5+\frac{5}5)-\frac{5}5=2019$$

Eight 6s (thanks @TheSimpliFire)

$$6\sqrt{6^6}+6!+\frac{6+6+6}6=2019$$

Ten 6s

$$(666+6+\frac{6}{6})\times\frac{6+6+6}{6}=2019$$

Ten 8s

$$\frac{\sqrt{8^8}}{\sqrt{\sqrt{8+8}}} - \sqrt{(8+8)}! - \sqrt{8+8} - \frac{8}{8} = 2019$$
$$\sqrt{8^8}=4096$$
$$\sqrt{\sqrt{8+8}}=2$$
$$\sqrt{(8+8)}!=24$$

• 8: $6\sqrt{6^6}+6!+\frac{6+6+6}6=2019$ Sep 12, 2019 at 14:11
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

Seven 4's no concatenation

\begin{align}\frac{\sqrt{\sqrt{4^{4!}}}}{\sqrt{4}} - 4! - 4 - \frac{4}{4}&= \frac{4096}{2} - 4! - 4 - \frac{4}{4}\\&= 2048 - 24 - 4 - 1\\&= 2019\end{align}

• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

I have found a solution with 7 (threes)

$$3!^3*3!+(3*3)^3-3! = 6^3*6*(9)^3-6=1296 + 729 - 6 = 2019$$

• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

All solutions without concatenation.

A solution for $$0$$ using 13 digits

$$(0! + 0! + 0!)!^{0! + 0! + 0! + 0!} + ((0! + 0! + 0!)!)! + 0! + 0! + 0! = 1296 + 720 + 3 = 2019$$

A solution for $$7$$ using 10 digits

$$\frac{(\frac{7! + 7 + 7}{7} - (7\times7)) \times (7+7+7)}{7} = \frac{673 * 21}{7} = 2019$$

A solution for $$8$$ using 8 digits

$$\dfrac{8!}{(\sqrt{8+8})!-\sqrt{8+8}}+\dfrac{(\sqrt{8+8})!}{8} = 2016 + 3 = 2019$$

Old solutions:

A solution for $$8$$ using 9 digits:

$$\frac{8! + 8*8 - \sqrt{8+8}}{8+8+\sqrt{8+8}} = \frac{40320 + 64 - 4}{16 + 4} = 2019$$

• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01
• It is interesting how each expression (except 7 and 8) is of the form $1296+720+3$ or $2048-29$. I wonder if these two can be proven to be optimal Sep 14, 2019 at 7:56

a "3-expression" instead of 5 in example:

$$((3!^3 \times 3)+((3! \times (3+ \frac 33))+ \frac 33)) \times 3$$
$$((6^3 \times 3) + ((6 \times 4)+1)) \times 3$$
$$((216 \times 3) + (24+1)) \times 3$$
$$(648+25) \times 3$$
$$673\times3 = 2019$$
10 threes

• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

A solution for 7 using 18 7's

$$7\times7\times7\times7 - \frac{777}{7} - \frac{777}{7} - \frac{777}{7} - 7\times7 = 2019$$
$$2401 - 111 -111 -111 -49$$

• I already submitted a solution of 16 $7$s, and @TheSimpliFire has a solution of 12. Sep 12, 2019 at 18:35
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01

12 1s:

$$(1+1)^{11} - (1+1+1)^{1+1+1} - (1+1)$$

• 11: $(1+1)^{11}-(1+1+1)(11-1)+1$ Sep 12, 2019 at 20:45
• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01
• @TheSimpliFire You're good at cutting that one extra digit off, haha. Sep 13, 2019 at 15:51

Nine 3s without concatenation.

$$3(3\times3)^3-3!\left(3^3+\dfrac{3}{3}\right)$$

• Based on the rules established by this meta post and the consensus around it, an answer must have justification for why the solution is optimal. Without that, this is a comment, not an answer.
– Deusovi
Sep 13, 2019 at 4:01
• @HerbWolfe $3(9^3)-6(27+1)=2019$ Sep 13, 2019 at 8:42

Nine ones with concatenation and decimal point:

$$(1+1)*\frac{1111}{1.1}-1$$

Eight fives with concatenation:

$$5 - \frac{5555}{5} + 5^5$$